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A124780
a(n) = gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.
4
1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 13, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 37, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1
OFFSET
0,2
COMMENTS
a(n) divides n+3 because A(n+2) = (n+2)(n+1)*A(n) + n+3.
LINKS
J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641.
J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
FORMULA
a(n) = gcd(A000522(n), A000522(n+2)) = (n+3)/A124782(n)
EXAMPLE
a(2) = gcd(A(2), A(4)) = gcd(5, 65) = 5.
MATHEMATICA
(A[n_] := Sum[n!/k!, {k, 0, n}]; Table[GCD[A[n], A[n+2]], {n, 0, 100}])
GCD[#[[1]], #[[3]]]&/@Partition[Table[Sum[n!/k!, {k, 0, n}], {n, 0, 100}], 3, 1] (* Harvey P. Dale, Jun 14 2022 *)
PROG
(PARI)
A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
A124780(n) = gcd(A000522(n), A000522(n+2)); \\ Antti Karttunen, Jul 07 2017
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Nov 07 2006
STATUS
approved